# Is this theorem already in existence?

I came up with this theorem (shown below) a few months ago, and I haven't been able to find anything like it on the web. This theorem will give you the quadratic expression that results from $(x-(a+bi))(x-(a-bi))$. This theorem is useful for finding all roots of an expression when given a complex root (because complex roots come in conjugate pairs, remember?). Can anyone tell me if this is old news, or if I have discovered something?

Some context:
I'm the student not the teacher. My teacher (which, BTW, she's my Mom, since I'm homeschooled) is trying to get me to re-prove every time.

Basic Theorem (which if it doesn't have a name, I'll call it Smith's Product of Complex Conjugate Roots Theorem. A tad long, don't you think?):

$(x-(a+bi))(x-(a-bi)) = x^2-2ax+a^2+b^2$

Proof 1:

$(x-(a+bi))(x-(a-bi))$

$(x-a-bi)(x-a+bi)$

$x^2-ax+bxi-ax+a^2-abi-bxi+abi-b^2i^2$

$x^2-2ax+a^2-b^2i^2$

$x^2-2ax+a^2+b^2$

Proof 2:

$(x-(a+bi))(x-(a-bi))$

$(x-a-bi)(x-a+bi)$

$((x-a)-bi)((x-a)+bi)$

$(x-a)^2-b^2i^2$

$(x^2-2ax+a^2)+b^2$

$x^2-2ax+a^2+b^2$

• Both proofs are fine, the second better. The result as far as I know does not have a name. – André Nicolas Jun 2 '15 at 22:47
• People have discovered how to multiply one degree-$1$ polynomial by another, unfortunately. – Nick D. Jun 2 '15 at 22:51
• First: congratulations on your independent thinking. Finding out something for the first time for yourself is valuable even if other people knew it. Second: it's important in your schooling in math (at home or anywhere) to find the right balance between rediscovering (and reproving) everything and building on known things in order to get to deeper mathematics faster. You and your Mom should work together on finding the right balance for you. – Ethan Bolker Jun 3 '15 at 0:06
• I'm happy to see a student excited by his or her own mathematical discovery. Keep on studying! There is an ocean of mathematics out there to explore. – suneater Jun 3 '15 at 0:14
• @ajotatxe. First, for any citizen of a country that is a signatory to the Berne Convention, setting your result down in tangible form (like posting here) automatically establishes your copyright. Second, whether or not it did, everything published here has a timestamp, establishing the primacy of the result (assuming that it is indeed new). While this wouldn't be the best place to publish a groundbreaking result, it would serve. – Rick Decker Jun 3 '15 at 0:30

While it should be encouraged that you reach your own results and ways of thinking, you should be more careful about the importance (?) of the results you achieve, other than to yourself.

Your question can have many interpretations, really: an excited student that tries to reach its own conclusions, an arrogant student, an independent student etc.

But, responding to your question: No, there is no name for this. This is a trivial consequence of basic properties of multiplication and complex numbers.

• Thanks. This really did answer the basic question. – Daniel Jun 2 '15 at 23:05

Your proof looks correct. However, this is highly unlikely to be a real theorem, because it is just simple binomial multiplication. This is similar to proving a theorem like "Addition Theorem" which states that 1+5=6. (Note I didn't use 1+1=2 because Peano and such.)

• @Daniel It is not like a proof, it is more similar to work because it is just binomial multiplication. For example, sure, we could make a theorem that says $(x_0+iy_0)(x_1+iy_1)=x_0x_1+i(x_0y_1+y_0x_1)-y_0y_1,$ but I would assume most students will just do the standard multiplication. – Teoc Jun 2 '15 at 22:49
• @Daniel Just a reminder that the polynomial requires real coefficients too. – JessicaK Jun 2 '15 at 22:51
• @Daniel Still, the number of steps required to actually multiply out the product using your second method is so low that the "naming" of a new "theorem" seems unnecessary at best, and probably just silly. – Zubin Mukerjee Jun 2 '15 at 22:52
• It seems that the question is not about the validity of your mathematics, but rather over who is "right' about how much work must be shown on a middle school algebra problem. Your proofs are nice. There of course reaches a point where real mathematicians accept basic fact as, well, fact, and the ability to prove such lemmas becomes meaningless on the whole. – Vladhagen Jun 2 '15 at 23:01
• @Daniel my point is that I don't write it out every time. I do such things in my head. Being able to prove things is good, as it teaches you how to prove other things, but eventually you just need to accept the basic rules of algebra as fact and omit the proofs. It's sort of like how you do not rewrite the Linux kernel every time you want to boot up your computer and run an operating system. Writing the Linux kernel each time shows prowess in computer science, but it also has a disturbing amount of inefficiency associated with it. – Vladhagen Jun 2 '15 at 23:08

It is a well-known fact (even in High School) that if $u$ and $v$ are the roots of a second degree, monic polynomial, then the polynomial is $$X^2-sX+p$$ where $s$ and $p$ are, respectively, $u+v$ and $uv$. Your "theorem" is a trivial consequence of this.

• I'm sorry, but that is far, far from a well known fact in high schools. Perhaps in some specialised mathematics gymnasiums, but in general, just, no. Most high-schoolers don't know what a polynomial is. – Davor Jun 3 '15 at 11:53
• Well, I teach that to my 14 years old pupils. And it is not a specialised gymnasium. – ajotatxe Jun 3 '15 at 13:39
• Certainly, polynomials are known to non-specialized high school students in my culture. The above "well-known" fact, however -- not so much. – Thomas Jun 3 '15 at 14:59
• I think it's usually introduced to high-schoolers when factoring quadratics with integer coefficients: you expand out the product and get them to find two integers which add up to $s$ and multiply to $p$. – Thomas Jun 3 '15 at 15:20
• You're all failing to make a difference between "introduced to" and "known to". – Davor Jun 3 '15 at 19:13

You get to the result much faster if you set $z=a+bi$, so $\bar{z}=a-bi$, the complex conjugate. Then $$(x-(a+bi))(x-(a-bi))=(x-z)(x-\bar{z})=x^2-(z+\bar{z})x+z\bar{z}$$ Since $z+\bar{z}=2a$ and $z\bar{z}=|z|^2=a^2+b^2$, the final form of the expression is $$x^2-2ax+a^2+b^2$$

You might note also that if you have a polynomial $x^2-2ax+q$ with negative discriminant, so $a^2-q<0$, you can set $b=\sqrt{q-a^2}$ and thus $q=a^2+b^2$. Then the polynomial is $$(x-z)(x-\bar{z})$$ where $z=a+bi$.

• Perhaps faster, but this is how I proved it off the top of my head when I came up with it. – Daniel Jun 2 '15 at 22:56
• @Daniel Very good, then! – egreg Jun 2 '15 at 23:00